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MANE 3332.05

Lecture 20

Agenda

  • Haven't started partial credit requests yet
  • Start Chapter 8
  • Chapter 8, Case 1 Practice Problems (assigned 11/6/2025, due 11/11/2025)
  • Attendance
  • Questions?

Handouts

Week Tuesday Lecture Thursday Lecture
10 11/4 - Return Midterm 11/6 - Chapter 8 (part 1)
11 11/11 - Chapter 8 (part 2) 11/13 - Chapter 8 (part 3)
12 11/18 - Chapter 8 (part 4) 11/25 - Chapter 9 (part 1)
13 11/25 - Chapter 9 (part 2) 11/27 - Thanksgiving Holiday (no class)
14 12/2 - Chapter 9 (part 3) 12/4 - Linear Regression
15 12/9 - Review Session 12/11 - Study Day (no class)

The final exam for MANE 3332.01 is Thursday December 18, 2025 at 1:15 PM - 3:00 PM.

Chapter 8 Introduction

  • Intervals are another method of performing estimation

  • A confidence interval is an interval estimate on a population parameter (primary focus of this chapter)

  • Three types of interval estimates

    • A confidence intervals bounds population or distribution parameters

    • A tolerance interval bounds a selected proportion of a distribution

    • A prediction interval bounds future observations from the population or distribution

  • Interval estimates, especially confidence intervals are commonly used in science and engineering

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Confidence Interval on the Mean of a normal distribution, variance known (Case 1)

  • Suppose that \(X_1,X_2,\ldots,X_n\) is a random sample from a normal population with unknown mean \(\mu\) and known variance \(\sigma^2\)

  • A general expression for a confidence interval is

\[ P\left[L\leq\mu\leq U\right]=1-\alpha \]
  • Using the sample results we calculate a \(100(1-\alpha)\%\) confidence of the form
\[ l\leq\mu\leq u \]
  • A \(100(1-\alpha)\%\) confidence interval for the mean of a normal distribution with variance known is
\[ \bar{x}-z_{\alpha/2}\frac{\sigma}{\sqrt{n}}\leq\mu\leq\bar{x}+z_{\alpha/2}\frac{\sigma}{\sqrt{n}} \]

where \(z_{\alpha/2}\) is the upper \(100\alpha/2\) percentage point of the standard normal distribution.

Problem 8-12,part a (6th edition)

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Interpreting Confidence Intervals

Montgomery gives the following statement regarding the correct interpretation of confidence intervals.

The correct interpretation lies in the realization that a CI is a random interval because in the probability statement defining the end-points of the interval, \(L\) and \(U\) are random variables. Consequently, the correct interpretation of a \(100(1-\alpha)\)% CI depends on the relative frequency view of probability. Specifically, if an infinite number of random samples are collected and a \(100(1-\alpha)\)% confidence interval for \(\mu\) is computed from each sample, \(100(1-\alpha)\)% of these intervals will contain the true value of \(\mu\).

One-sided Confidence Bounds

It is possible to construct on-sided confidence bounds

  • A \(100(1-\alpha)\%\) upper-confidence bound for \(\mu\) is
\[ \mu\leq u=\bar{x}+z_\alpha\frac{\sigma}{\sqrt{n}} \]
  • A \(100(1-\alpha)\%\) lower-confidence bound for \(\mu\) is
\[ \bar{x}-z_\alpha\frac{\sigma}{\sqrt{n}}=l\leq\mu \]

Sample Size Considerations

If \(\bar{x}\) is used as an estimate of \(\mu\), we can be \(100(1-\alpha)\%\) confident that the error \(|\bar{x}-\mu|\) will not exceed a specified amount \(E\) when the sample size is

\[ n=\left(\frac{z_{\alpha/2}\sigma}{E}\right)^2 \]

Problem 8--12,part b (6th edition)

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A Large Sample CI for \(\mu\)

  • When \(n\) is large (say greater than or equal to 40), the central limit theorem can be used

  • It states that \(\frac{\bar{x}-\mu}{\sigma/\sqrt{n}}\) is approximately a standard normal random variable.

  • Thus, we can replace the quantity \(\sigma/\sqrt{n}\) with \(S/\sqrt{n}\) and still use the quantiles of the normal distribution to construct a confidence interval.

\[ \bar{x}-z_{\alpha/2}\frac{s}{\sqrt{n}}\leq\mu\leq\bar{x}+z_{\alpha/2}\frac{s}{\sqrt{n}} \]
  • What assumption did we relax and why?

Chapter 8, Case 1 Practice Problems

Confidence Interval for the mean of Normal distribution with variance unknown (Case 2)

The \(t\) distribution

  • Definition.

Let \(X_1, X_2,\ldots,X_n\) be a random sample from a normal distribution with unknown mean \(\mu\) and unknown variance \(\sigma^2\). The random variable

\[ T=\frac{\overline{X}-\mu}{S/\sqrt{n}} \]

has a \(t\) distribution with \(n-1\) degrees of freedom.

  • A table of percentage points (quantiles) is given in Appendix A Table 5

  • Figure 8-4 on page 180 shows the relationship between the \(t\) and normal distributions.

  • Figure 8-5 explains the percentage points of the \(t\) distribution

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Confidence interval definition

  • Using the \(t\) distribution it is possible to construct CIs

If \(\bar{x}\) and \(s\) are the mean and standard deviation of a random sample from a normal distribution with unknown variance \(\sigma^2\), a \(100(1-\alpha)\%\) confidence interval on \(\mu\) is given by

\[ \bar{x}-t_{\alpha/2,n-1}\frac{s}{\sqrt{n}}\leq\mu\leq \bar{x}+t_{\alpha/2,n-1}\frac{s}{\sqrt{n}} \]

where \(t_{\alpha/2,n-1}\) is the upper \(100(\alpha/2)\) percentage point of the \(t\) distribution with \(n-1\) degrees of freedom.

Problem 8-30 (6th edition)

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One-sided confidence bounds

  • Are easy to construct

  • Use only the appropriate upper or lower bound

  • Change \(t_{\alpha/2,n-1}\) to \(t_{\alpha,n-1}\)

Chapter 8, Case 2 Practice Problems

Confidence Interval for \(\sigma^2\) and \(\sigma\) (Case 3)

  • Section 8-3 presents a CI for \(\sigma^2\) or \(\sigma\)

  • Requires the \(\chi^2\) (chi-squared) distribution

Let \(X_1,X_2,\ldots,X_n\) be a random sample from a normal distribution with mean \(\mu\) and variance \(\sigma^2\) and let \(S^2\) be the sample variance. Then the random variable

\[ \chi^2=\frac{(n-1)S^2}{\sigma^2} \]

has a chi-square (\(\chi^2\)) distribution with \(n-1\) degrees of freedom

  • A table of the upper percentage points of the \(\chi^2\) distribution are given in Table 4 in the appendix

  • Figure 8-9 on page 183 explains the percentage points of the \(\chi^2\) distribution

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Confidence Intervals for \(\sigma^2\) and \(\sigma\)

If \(s^2\) is the sample variance from a random sample of \(n\) observations from a normal distribution with unknown variance \(\sigma^2\), then a \(100(1-\alpha)\%\) confidence interval on \(\sigma^2\) is

\[ \frac{(n-1)s^2}{\chi_{\alpha/2,n-1}^2}\leq\sigma^2\leq\frac{(n-1)s^2}{\chi_{1-\alpha/2,n-1}^2} \]

where \(\chi^2_{\alpha/2,n-1}\) and \(\chi^2_{1-\alpha/2,n-1}\) are the upper and lower \(100\alpha/2\) percentage points of the \(\chi^2\)-distribution with \(n-1\) degrees of freedom

Problem 8-36 (6th edition)

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One-sided confidence bounds

  • Are easy to construct

  • Use only the appropriate upper or lower bound

  • Change \(\chi^2_{\alpha/2,n-1}\) to \(\chi^2_{\alpha,n-1}\) or \(\chi^2_{1-\alpha/2,n-1}\) to \(\chi^2_{1-\alpha,n-1}\)

  • See eqn (8-20) on page 184

Chapter 8, Case 3 Practice Problems

Large-Sample CI for a Population Proportion (Case 4)

  • Recall from chapter 4, that the sampling distribution of \(\widehat{P}\) is approximately normal with mean \(p\) and variance \(p(1-p)/p\), if \(n\) is not too close to either 0 or 1 and if \(n\) is relatively large.

  • Typically, we require both \(np\geq 5\) and \(n(1-p)\geq 5\)

f \(n\) is large, the distribution of

\[ Z=\frac{X-np}{\sqrt{np(1-p)}}=\frac{\widehat{P}-p}{\sqrt{\frac{p(1-p)}{n}}} \]

is approximately standard normal.

If \(\hat{p}\) is the proportion of observations in a random sample of size \(n\) that belongs to a class of interest, an approximate \(100(1-\alpha)\%\) confidence interval on the proportion \(p\) of the population that belongs to this class is

\[ \hat{p}-z_{\alpha/2}\sqrt{\frac{\hat{p}(1-\hat{p})}{n}}\leq p\leq \hat{p}+z_{\alpha/2}\sqrt{\frac{\hat{p}(1-\hat{p})}{n}} \]

where \(z_{\alpha/2}\) is the upper \(\alpha/2\) percentage point of the standard normal distribution

Other Considerations

  • We can select a sample so that we are \(100(1-\alpha)\%\) confident that error \(E=|p-\widehat{P}|\) using
\[ n=\left(\frac{z_{\alpha/2}}{E}\right)^2p(1-p) \]
  • An upper bound on is given by
\[ n=\left(\frac{z_{\alpha/2}}{E}\right)^2(0.25) \]
  • One-sided confidence bounds are given in eqn (8-26) on page 187

Guidelines for Constructing Confidence Intervals

  • Review excellent guide given in Table 8-1

Other Interval Estimates

  • When we want to predict the value of a single value in the future, a prediction interval is used

  • A tolerance interval captures \(100(1-\alpha)\%\) of observations from a distribution

Prediction Interval for a Normal Distribution

  • Excellent discussion on pages 189 - 190

  • A \(100(1-\alpha)\%\) PI on a single future observation from a normal distribution is given by

\[ \bar{x}-t_{\alpha/2,n-1}s\sqrt{1+\frac{1}{n}}\leq X_{n+1}\leq \bar{x}+t_{\alpha/2,n-1}s\sqrt{1+\frac{1}{n}} \]

Tolerance Intervals for a Normal Distribution

  • A tolerance interval to contain at least \(\gamma\%\) of the values in a normal population with confidence level \(100(1-\alpha)\%\) is
\[ \bar{x}-ks,\bar{x}+ks \]

where \(k\) is a tolerance interval factor for the normal distribution found in appendix A Table XII. Values are given for \(1-\alpha\)=0.9, 0.95 and 0.99 confidence levels and for \(\gamma=.90,\,.95,\,\mbox{and }.99\%\) probability of coverage

  • One-sided tolerance bounds can also be computed. The factors are also in Table XII